Ncert Solutions Maths class 12th

$\sim \left(\sim p\wedge \left(p\vee q\right)\right)=pV\left(\sim p\wedge \sim q\right)$

$\begin{array}{l}=\left(pv\sim p\right)\wedge \left(pv\sim q\right)\\ =pv\sim q\end{array}$

 $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill \alpha \hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -\alpha \hfill & \hfill 1\hfill \end{array}\right|=\left(4\alpha \widehat{i}+8\widehat{j}-4\alpha \widehat{k}\right)$

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt[]{32{\alpha }^2+64}=8\sqrt[]{3}$

322 + 64 = 192

2 = $\frac{128}{32}=4$

$\overrightarrow{a}.\overrightarrow{b}=3-{\alpha }^2+3=6-{\alpha }^2=6-4=2$

Area of shaded region

$2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(2x^2+9-5x^2\right)dx}$

$=2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(9-3x^2\right)dx}$

${\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(3-x^2\right)dx=6{\left[3x-\frac{x^3}{3}\right]}_0^{\sqrt[]{3}}=6\left[\frac{9\sqrt[]{3}-3\sqrt[]{3}}{3}\right]}=12\sqrt[]{3}$

 $f\left(x\right)=\left\{\begin{array}{cc}\hfill min\left\{\left|x\right|,2-x^2\right\},\hfill & \hfill -2\le x\le 2\hfill \\ \hfill \left[\left|x\right|\right],\hfill & \hfill 2\le \left|x\right|\le 3\hfill \end{array}\right\}$

Number of points where f is not differentiable = 5

$I=3{\displaystyle \underset{-2}{\overset{2}{\int }}\left|x^2-x-2\right|dx}$

$=3\left({\displaystyle \underset{-2}{\overset{-1}{\int }}\left(x^2-x-2\right)dx-{\displaystyle \underset{-1}{\overset{2}{\int }}\left(x^2-x-2\right)dx}}\right)$

$=3\left[\left(\frac{7}{6}+\frac{2}{3}\right)-\left\{-\frac{10}{3}-\frac{7}{6}\right\}\right]$

$=3\left(\frac{11}{6}+\frac{27}{6}\right)=\frac{38}{2}=19$

Total possibilities = 2⁵ × 2⁵

Farounable case = ⁵C₂ × 3³ = 10 × 3³

$\therefore requiredprobability=\frac{10\times 3^3}{2^5\times 2^5}=\frac{5\times 27}{2^9}=\frac{135}{2^9}$

$2x{y}^2-y)dx+xdy=0$

$\Rightarrow \frac{dy}{dx}+2y^2-\frac{y}{x}=0$

$Put\frac{1}{y}=z$

Then $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$

$\frac{dz}{dx}+\frac{1}{x}z=2$

Point (2, 1) c = 2 – 4 = 2 y = $\frac{x}{x^2-2}$

$\left|y\left(1\right)\right|=1$

${\mathcal{l}}_1:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$

${\mathcal{l}}_2:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=\left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}:\overrightarrow{r}=\overrightarrow{0}+\lambda \left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}\&{\mathcal{l}}_1\to$

Point of intersection $\mathcal{l}\&{\mathcal{l}}_1$ is P (2, 3, 2)

Point Q on ${\mathcal{l}}_2$ is (3 + 2s, 3 + 2s, 2 + s).

  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 1

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$